Answer to Question #223828 in Chemical Engineering for Lokika

Question #223828

The differential equation of (x-a)^{2}+(y-b)^{2}+z^{2}=21


1
Expert's answer
2021-08-16T02:20:56-0400
z=(xa)2+(yb)2z=(x-a)^2+(y-b)^2

Differentiating partially with respect to xx and y,y, we get



zx=2(xa)\dfrac{\partial z}{\partial x}=2(x-a)zy=2(yb)\dfrac{\partial z}{\partial y}=2(y-b)

Squaring and adding these equations, we have



(zx)2+(zy)2=(2(xa))2+(2(yb))2(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=(2(x-a))^2+(2(y-b))^2(zx)2+(zy)2=4((xa)2+(yb)2)(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=4((x-a)^2+(y-b)^2)

Since (xa)2+(yb)2=z,(x-a)^2+(y-b)^2=z, we get



(zx)2+(zy)2=4z(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=4z

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment